Question

A coffee container is $\dfrac{3}{5}$ full of beans. These beans are then put into another container having volume thrice of the first one. What fraction of the large box is filled with beans?
(A) $\dfrac{2}{5}$
(B) $\dfrac{1}{5}$
(C) $\dfrac{3}{5}$
(D) $1$

Answer 1

Hint: Start with assuming a variable for the volume of the initial container as $'m'$ . Then according to the fraction, the volume of the bean will become $\dfrac{3}{5}m$ . Now use the definition of the fraction to calculate the fraction of the large box filled with beans by dividing the volume of the beans by the volume of the larger container.

Complete step-by-step answer:
Here in this problem, we are given a container which is $\dfrac{3}{5}$ filled with beans. These beans are then emptied into a container having the volume thrice as the volume of the previous container. Using this information we need to find what fraction of the larger box is filled with box.
For representing the information we can assume the volume of the initial container as some variable. Let the volume of the smaller container be $'m'$
Therefore, the volume of larger container $ = $ $3$ times the volume of the smaller container $ = 3 \times m = 3m$
Since three-fifth of the smaller container is filled with beans, the volume of the beans can be expressed as $ = \dfrac{3}{5} \times m = \dfrac{3}{5}m$
So now we have the volume of beans and the volume of the new container in terms of the volume of the smaller container.
A fraction can be defined as the ratio of a part from the whole and here we have the volume of beans as a part and the volume of the larger container as the whole.
Therefore, the required fraction can be represented by:
$ \Rightarrow \dfrac{3}{5}m \div 3m = \dfrac{{\dfrac{3}{5}m}}{{3m}}$
This can be solved to obtain the required fraction
$ \Rightarrow \dfrac{{\dfrac{3}{5}m}}{{3m}} = \dfrac{{3m}}{{5 \times 3m}}$
After dividing numerator and denominator by $3m$ we get:
$ \Rightarrow \dfrac{{\dfrac{3}{5}m}}{{3m}} = \dfrac{{3m}}{{5 \times 3m}} = \dfrac{1}{5}$
Therefore, we get the required fraction of a larger box filled with beans as $\dfrac{1}{5}$ .
Hence, the option (B) is the correct answer.

Note: In questions like this utilizing the basic definition of fractions always play an important role. An alternative approach can be to assume the volume of beans as $'3n'$ and the volume of the initial container as $'5n'$ according to the given fraction. Then we get the volume of the larger container as $3 \times 5n = 15n$ . Then the required fraction can be calculated by solving $\dfrac{{3n}}{{15n}} = \dfrac{1}{5}$ .